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rusak2 [61]
3 years ago
14

A car of mass 600kg is moving at 15m/s. The driver accelerate gently to a final velocity of 30m/s so that a force of force acts

on the car for 10s. Calculate the magnitude of the force that acting on the car during the 10s time interval.
Physics
2 answers:
Rudik [331]3 years ago
8 0

Answer:

<h2>F = 900 N</h2>

Explanation:

m = 600 kg

vo = 15 m/s

vt = 30m/s

t = 10 s

F = ?

vt = vo + at

30 = 15 + a (10)

15 = 10a

a = 15/10 = 1.5 m/s²

F = m. a

F = 600 . 1.5

<h2>F = 900 N</h2>

Anvisha [2.4K]3 years ago
3 0

Answer:

Really hope you get it!!

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Gamma rays x rays visible light and radio waves are all types of
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Answer:

Electromagnetic waves

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and they are classified into 7 different types, according to their frequency. From lowest to highest frequency, we have:

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Doctor prescribed drug which is 500mg for patient change this unit into gram
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Answer:

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6 0
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Two 110 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Z at –10 m/s when t
salantis [7]

Explanation:

Mass of bumper cars, m_1=m_2=110\ kg

Initial speed of car A, u_1=8\ m/s

Initial speed of car Z, u_2=-10\ m/s

Final speed of car A after the collision, v_1=-10\ m/s

We need to find the velocity of car Z after the collision. Let it is equal to v_2. Using the conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

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v_2=\dfrac{-1320}{110}\ m/s

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5 0
3 years ago
A 2kg hockey puck is sliding across the ice skating rink at 2 m/s. A player hits the puck so it's velocity increases to 10 m/s.
konstantin123 [22]

The work done on the puck is 96 J

Explanation:

According to the work-energy theorem, the work done on the hockey puck is equal to the change in kinetic energy of the puck.

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where

K_f = \frac{1}{2}mv^2 is the final kinetic energy of the puck, with

m = 2 kg being the mass of the puck

v = 10 m/s is the final speed

K_i = \frac{1}{2}mu^2 is the initial kinetic energy of the puck, with

u = 2 m/s being the initial speed of the puck

Substituting numbers into the equation, we find the work done by the player on the puck:

W=\frac{1}{2}(2)(10)^2 - \frac{1}{2}(2)(2)^2=96 J

Learn more about work and kinetic energy:

brainly.com/question/6763771  

brainly.com/question/6443626  

brainly.com/question/6536722

#LearnwithBrainly

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